Problem: Which of the following numbers is a multiple of 8? ${74,99,112,114,118}$
Answer: The multiples of $8$ are $8$ $16$ $24$ $32$ ..... In general, any number that leaves no remainder when divided by $8$ is considered a multiple of $8$ We can start by dividing each of our answer choices by $8$ $74 \div 8 = 9\text{ R }2$ $99 \div 8 = 12\text{ R }3$ $112 \div 8 = 14$ $114 \div 8 = 14\text{ R }2$ $118 \div 8 = 14\text{ R }6$ The only answer choice that leaves no remainder after the division is $112$ $ 14$ $8$ $112$ We can check our answer by looking at the prime factorization of both numbers. Notice that the prime factors of $8$ are contained within the prime factors of $112$ $112 = 2\times2\times2\times2\times7 8 = 2\times2\times2$ Therefore the only multiple of $8$ out of our choices is $112$. We can say that $112$ is divisible by $8$.